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A Banach lattice $E$ has the

$(1)$ Schur property provided that any weakly null sequence in $E$ is norm null in $E$, and

$(2)$ positive Schur property provided that any weakly null sequence of positive elements in $E$ is norm null.

$L_1$ is an easy example of a lattice which has $(2)$ and not $(1)$. Indeed, since the functional $f\mapsto \int_0^1 f$ is norming for all positive functions in $L_1$, if $(f_n)_{n=1}^\infty\subset L_1^+$ is weakly null, $$\|f_n\|_{L_1}=\int_0^1 f_n \to 0.$$ So $L_1$ has the positive Schur property. However, $L_1$ contains copies of $\ell_2$, the canonical basis of which is weakly null and not norm null, so $L_1$ fails the Schur property.

There are other examples. If $E$ is a Banach lattice with the Schur property whose lattice structure comes from a $1$-unconditional basis, then we may take the direct sum $(\oplus L_1)_E$. This has the positive Schur property (which isn't too hard to see), and fails the Schur property because it contains a copy of $L_1$.

These two examples are mentioned in a paper of Wnuk, "Some characterizations of Banach lattices with the Schur property."

In these examples, the space failing the Schur property can be seen by a normalized, weakly null sequence whose Cesaro means converge in norm to zero (which, in some sense, means the sequence is "very" weakly null).

Are there other examples of Banach lattices which enjoy the positive Schur property, fail the Schur property, but still do not contain any normalized, weakly null sequence all of whose subsequences have Cesaro means converging to zero in norm? By a well known dichotomy of Rosenthal, this is equivalent to: $E$ has the positive Schur property, fails the Schur property, and every seminormalized, weakly null sequence in $E$ has a subsequence which generates a spreading model isomorphic to $\ell_1$.

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